Problem Statement: A drop of water with a mass of 10^-6 kg and a charge of -10^6 C is subjected to an electric field. What is the magnitude and direction of the electric field required to balance the weight of the drop in each of the following cases: 1) 10 V/m upward, 2) 10 V/m downward, 3) 0.1 V/m downward, 4) 0.1 V/m upward?

Solution:
To solve this problem, we need to use the following equation:
F = qE
where F is the electrical force acting on the drop, q is the charge on the drop, and E is the electric field.

We can also use the equation for weight:
W = mg
where W is the weight of the drop, m is its mass, and g is the acceleration due to gravity.

If the drop is in a balanced condition, the electrical force must be equal and opposite to the weight of the drop. Therefore, we can set F = W and solve for E.

Case 1: 10 V/m upward
In this case, the electric field is acting in the upward direction, so the electrical force on the drop is also upward. To balance the weight of the drop, the electric field should be downward. Let's calculate the magnitude of the electric field required to balance the weight of the drop:
F = qE
W = F
mg = qE
E = mg/q
E = (10^-6 kg)(9.8 m/s^2)/(-10^6 C)
E = -9.8 V/m

Therefore, the magnitude of the electric field required to balance the weight of the drop is 9.8 V/m downward.

Case 2: 10 V/m downward
In this case, the electric field is acting in the downward direction, so the electrical force on the drop is also downward. To balance the weight of the drop, the electric field should be upward. Let's calculate the magnitude of the electric field required to balance the weight of the drop:
F = qE
W = F
mg = qE
E = mg/q
E = (10^-6 kg)(9.8 m/s^2)/(-10^6 C)
E = -9.8 V/m

Therefore, the magnitude of the electric field required to balance the weight of the drop is 9.8 V/m upward.

Case 3: 0.1 V/m downward
In this case, the electric field is acting in the downward direction, so the electrical force on the drop is also downward. To balance the weight of the drop, the electric field should be upward. Let's calculate the magnitude of the electric field required to balance the weight of the drop:
F = qE
W = F
mg = qE
E = mg/q
E = (10^-6 kg)(9.8 m/s^2)/(-10^6 C)
E = -9.8 V/m

Since the given electric field is already in the downward direction, we need to subtract it from the magnitude of the electric field required to balance the weight of the drop:
E_required = -9.8 V/m
E_balance = E_required - 0.1 V/m
E_balance

We know that, F = qE . As condition says that electric field is applied in such a way that is balanced. Moreover, charge is negative, therefore ,it will move away from electric field's direction.
F=mg = q E
10^-6×10 =10^-6 ×E

E = 10V/m downward.